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When photons of energy `4.25 eV` strike the surface of metal A, the ejected photoelectrons have maximum kinetic energy `T_(A)` eV and De-broglie wavelength `lambda_(A)`. The maximum energy of photoelectron liberated from another metal B by photon of energy 4.70 eV is `T_(B) = (T_(A) - 1.50) eV` if the de Brogle wavelength of these photoelectrons is `lambda_(B) = 2 lambda_(A) `, thenA. the work function of A is 2.25 eVB. the work funtion of B is 4.20 eVC. `T_A=2.00eV`D. `T_B=2.75eV` |
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Answer» Correct Answer - A::B::C For metal `A:4.25=W_A+T_A` Also, `T_A=(1)/(2)mv_A^2=(1)/(2)(m^2v_A^2)/(m)=(p_A^2)/(2m)(h^2)/(2mlamda_A^2)` For metal `B:4.7=(T_A-1.5)+W_B` Also, `T_B=(h^2)/(2mlamda_B^2)xx(2mlamda_A^2)/(h^2)=(lamda_A^2)/(lamda_B^2)` `implies(T_A-1.5)/(T_A)=(lamda_A^2)/(2lamda_A^2)=(lamda_A^2)/(4lamda_A^2)=(1)/(4)` `implies4T_A-6=T_AimpliesT_A=2eV` |
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